Let X denote the number of odds, X = 0, 1, 2, 3 The experiment follows Binomial distribution
\(n=3, p=\frac{1}{2}, q=\frac{1}{2}\)
The required probability = 1 – P(X = O)
\(=1-{ }^{3} C_{0}\left(\frac{1}{2}\right)^{3}\left(\frac{1}{2}\right)^{0}=1-\frac{1}{8}=\frac{7}{8}\)
(b) P(Two cards are aces with replacement) = \(\frac{4 \times 4}{52 \times 52}=\frac{1}{169}\)
We know there are 4 aces in a deck of 52 cards. Let X denote the number of aces. Then X can take values 0,1,2.
P(X0) = P(no ace and no ace) = P(no ace) x P(no ace)
\(=\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}\)
P(X=1)= P(ace and no ace or no ace and ace) = P(ace and no ace ) + P(no ace and ace)
\(=\frac{4}{52} \times \frac{48}{52}+\frac{48}{52} \times \frac{4}{52}=\frac{24}{169}\)
P(X =2) = P(ace & ace) = P(ace) x P(ace)
\(=\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}\)
| X |
0 |
1 |
2 |
| P(X) |
144/169 |
24/169 |
1/169 |
